posted by anonymous 78 on .
When nitrogen dioxide (NO2) from car exhaust combines w/ water in the air it forms nitric acid (HNO3), which causes acid rain, and nitrogen monoxide. 3NO2(g)+ H2O(L)----->2HNO3(aq)+ N(g) a.How many molecules of NO2 are needed to react with 0.300 mole H2O?
b. how many grams of HNO3 can be produced if 175g NO2 is mixed with 55.2g of H2O? the first one i posted is mixed up with the other problem from my homework thanks
You made a typo in the equation.
3NO2 + H2O ==> 2HNO3 + NO
0.300 mol H2O x (3 mols NO2/1 mol H2O) - 0.300 x 3/1 = 0.9 mol NO2 needed.
You know 1 mol NO2 has 6.02E23 molecules; you need to calculate the numb err in 0.9 mol.
b. This is a limiting reagent. I work them this way.
mols NO2 = grams/molar mass = ?
mols H2O = grams/molar mass = ?
Using the coefficients in the balanced equation, convert mols NO2 to mols HNO3.
Do the same to convert mols H2O to mols HNO3.
It is quite likely these two numbers for mols HNO3 will not be the same which means one of them must be wrong. The correct answer in limiting reagent problems is ALWAYS the smaller value and the reagent providing that number is the limiting reagent.
Convert the smaller number to grams. g = mols x molar mass.
how will I get the mass of HNO3 produced if 175g NO2 is mixed with 55.2 g H2O. Should I add both the grams of NO2 and H2O
No. The instructions for that part of the problem I gave under part b. Find mold NO2 and mols H2O, convert each to mols HNO3, use the SMALLER number of mols and convert to grams HNO3.
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